Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $q = \dfrac{3z^2 - 15z}{z^2 - 11z + 30} \times \dfrac{-5z^2 - 55z - 90}{-3z^2 - 27z} $
Solution: First factor out any common factors. $q = \dfrac{3z(z - 5)}{z^2 - 11z + 30} \times \dfrac{-5(z^2 + 11z + 18)}{-3z(z + 9)} $ Then factor the quadratic expressions. $q = \dfrac {3z(z - 5)} {(z - 5)(z - 6)} \times \dfrac {-5(z + 9)(z + 2)} {-3z(z + 9)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {3z(z - 5) \times -5(z + 9)(z + 2) } { (z - 5)(z - 6) \times -3z(z + 9)} $ $q = \dfrac {-15z(z + 9)(z + 2)(z - 5)} {-3z(z - 5)(z - 6)(z + 9)} $ Notice that $(z - 5)$ and $(z + 9)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {-15z(z + 9)(z + 2)\cancel{(z - 5)}} {-3z\cancel{(z - 5)}(z - 6)(z + 9)} $ We are dividing by $z - 5$ , so $z - 5 \neq 0$ Therefore, $z \neq 5$ $q = \dfrac {-15z\cancel{(z + 9)}(z + 2)\cancel{(z - 5)}} {-3z\cancel{(z - 5)}(z - 6)\cancel{(z + 9)}} $ We are dividing by $z + 9$ , so $z + 9 \neq 0$ Therefore, $z \neq -9$ $q = \dfrac {-15z(z + 2)} {-3z(z - 6)} $ $ q = \dfrac{5(z + 2)}{z - 6}; z \neq 5; z \neq -9 $